Question

A capacitor of capacitance C is initially charged to a potential difference of V volt. Now it is connected to a battery of 2V volt with opposite polarity. The ratio of final energy stored in the capacitor to heat generated will be

• A. 1/2
• B. 4/9
• C. 9/4
• D. 2

Answer 1

Answer: (B)

4/9

Answer 2

Here's how to solve this problem:

1. Initial Energy Stored:

   $U_i = \frac{1}{2}CV^2$

2. Final Energy Stored:

   $U_f = \frac{1}{2}C(2V)^2 = 2CV^2$

3. Charge Flow:

   Since the capacitor is connected with opposite polarity, the total charge that flows through the battery is:

   $\Delta q = |2CV - (-CV)| = 3CV$

4. Work Done by Battery:

   $W_{batt} = (3CV)(2V) = 6CV^2$

5. Heat Generated:

   Using energy conservation:

   $H = W_{batt} - (U_f - U_i) = 6CV^2 - (2CV^2 - \frac{1}{2}CV^2) = \frac{9}{2}CV^2$

6. Ratio:

   $\frac{U_f}{H} = \frac{2CV^2}{\frac{9}{2}CV^2} = \frac{4}{9}$

Therefore, the ratio of final energy stored to heat generated is $\frac{4}{9}$.